Transformations that preserve beta random variables Consider a random variable $x$ distributed according to a beta distribution $\text{Beta}(\alpha,\beta)$.
Now draw a sample $x\sim\text{Beta}(\alpha,\beta)$, and transform this sample via some function $f$, i.e., $y=f(x)$. Note that $f$ could be stochastic in that it represents a conditional distribution $p(y\mid x)$.
What conditions on $f$ ensure that $y$ is also distributed according to some beta distribution, $\text{Beta}(\alpha',\beta')$?
 A: Deterministic case
Notation: $F$ is the cdf of $\text{Beta}(\alpha,\beta)$ and $G$ is the cdf of $\text{Beta}(\alpha',\beta')$. Note that $\alpha,\beta$ are known and fixed. Observe that $F$ and $G$ are strictly increasing bijections $(0,1) \to (0,1)$.
Let $u := F(x)$ and $v := G(y)$. Suppose we have $y=f(x)$ with a deterministic function $f$. Now define
$$
h : (0,1) \to (0,1) : h(u) := G(f(F^{-1}(u))) = v.
$$
We know that $u \sim U(0,1)$. On the other hand, $y \sim G$ is equivalent to $v \sim U(0,1)$, which is equivalent to saying that $h$ maps a $U(0,1)$ random variable to a (possibly different) $U(0,1)$ random variable.
So a sufficient and necessary condition for $y \sim G$ is that $f = G^{-1} \circ h \circ F$ with any mapping $h: (0,1) \to (0,1)$ that preserves the measures of all Borel sets.
Now there are quite many such mappings. Some easy examples are $h(u)=u$ and $h(u)=1-u$, but more generally you could, for example, divide $(0,1)$ to $r$ intervals of length $1/r$, and $h$ could shuffle and mirror these intervals arbitrarily, like
$$
h(u) = \begin{cases}
 u+1/3 & \text{if $0<u \le 1/3$}\\
 u-1/3 & \text{if $1/3 < u \le 2/3$}\\
 u & \text{if $2/3 < u < 1$.}
 \end{cases}
$$
(Also, $h$ could map some zero-measure set wherever you want, like $h(u)=u^7/\pi$ whenever $u \in \mathbb Q$.) This is a pretty wild collection of functions and I'm not sure this is what the OP wants.
Deterministic and monotonous
Things get much more orderly if we impose an extra requirement that $f$ is monotonous (or that it is continuous). Then also $h$ must be monotonous and the only possibilities are $h(u)=u$ and $h(u)=1-u$.
If we want $f$ to be increasing, the only possibility is $h(u)=u$ and thus $f = G^{-1} \circ F$, where $G$ is the cdf of some Beta distribution. I guess this is the neatest condition you can get, if you want sufficient and necessary, because the inverse cdf's of Beta distributions do not in general have closed form expressions.
For just sufficient conditions, hmm ... if $x \sim \text{Beta}(\alpha,1)$ then $F(x) = x^\alpha$. Then $f(x)=x^r$ with $r>0$ is a sufficient condition, because then $y \sim \text{Beta}(\alpha/r,1)$. But this is a very restricted family of beta distributions so probably not what is wanted.
For just necessary conditions ... Empirically, by plotting $f=G^{-1} \circ F$ for various Beta cdf functions $F,G$, it seems that $f$ has at most one inflection point in the interval $(0,1)$. Perhaps this would not be too hard to prove. But this is a relatively weak condition and nothing like sufficient.

Stochastic case
We have $x$, which has some beta distribution with cdf $F$, and we are looking for a conditional distribution $p(y|x)$ such that $y$ has a beta distribution with cdf $G$. If we again transform $u=F(x)$ and $v=G(y)$, we are looking for a conditional distribution $p(v|u)$, or equivalently a joint distribution $p(u,v)=p(v|u)p(u)$ where $u,v$ are uniform on $(0,1)$. Now such distributions are plenty: $p(u,v)$ can be any bivariate copula. So here, without further restrictions, the possibilities are even wilder than in the deterministic case.
